Ultra-QuickSort(归并排序)

http://poj.org/problem?id=2299

此题即为归并排序求逆序数。

#include<stdio.h>
#include<string.h>
const int N=500002;
int a[N],temp[N];
long long  ans;
void merge_arr(int first,int mid,int last)//合并区间
{
    int k = 0;
    int i = first,n = mid;
    int j = mid+1,m = last;
    while(i <= n && j <= m)
    {
        if (a[i] <= a[j])
            temp[k++] = a[i++];
        else
        {
            temp[k++] = a[j++];
            ans += (mid-i+1);//计算逆序数
        }
    }
    while(i <= n)
        temp[k++] = a[i++];
    while(j <= m)
        temp[k++] = a[j++];
    for (i = 0; i < k; i ++)
        a[first+i] = temp[i];//将有序的temp[]赋给原数组
}
void mergesort(int first,int last)//归并排序
{
    if (first < last)
    {
        int mid = (first+last)/2;
        mergesort(first,mid);
        mergesort(mid+1,last);
        merge_arr(first,mid,last);
    }
}
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        ans = 0;
        for (int i = 0; i < n; i ++)
        {
            scanf("%d",&a[i]);
        }
        mergesort(0,n-1);
        printf("%lld
",ans);
    }
    return 0;
}

同类型的题：

http://poj.org/problem?id=1804 

Brainman

题意：给出n个数，只能相邻的两个数交换，问最少交换多少次才能使这n个数递增排列。

思路：求出该序列的逆序数即为最少交换的次数。

法一：直接暴力o(n^2)

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int N=1005;

int a[N];

int main()
{
    int t,o = 0;
    scanf("%d",&t);
    while(t--)
    {
        o++;
        int n;
        scanf("%d",&n);
        for (int i = 0; i < n; i++)
            scanf("%d",&a[i]);

        int cnt = 0;
        for (int i = 0; i < n; i++)
        {
            for (int j = i+1; j < n; j++)
            {
                if (a[i] > a[j])
                    cnt++;
            }

        }
        printf("Scenario #%d:
%d

",o,cnt);
    }
    return 0;
}

法二：归并排序 o(nlogn)

#include <stdio.h>
#include <string.h>
const int N=1002;
const int INF = 1<<28;
long long ans = 0;
void Merge(int *a,int s,int mid,int t)
{
    int lenl= mid-s+1;
    int lenr = t-mid;
    int *b = new int[lenl+2];
    int *c = new int[lenr+2];
    for (int i = 1; i <= lenl; i++)
        b[i] = a[s+i-1];
    b[lenl+1] = INF;
    for (int i = 1; i <= lenr; i++)
        c[i] = a[mid+i];
    c[lenr+1] = INF;
    int i = 1 ,j = 1,k = s;
    while(k <= t)
    {
        if (b[i] <= c[j])
            a[k++] = b[i++];
        else
        {
            a[k++] = c[j++];
            ans+=lenl-i+1; //计算逆序数
        }
    }
    delete(b);
    delete(c);
}
void MergeSort(int *a,int s,int t)
{
    int mid = (s+t)>>1;
    if(t <= s)
        return ;
    MergeSort(a,s,mid);
    MergeSort(a,mid+1,t);
    Merge(a,s,mid,t);
}
int main()
{
    int t,o = 0,a[N],b[N];
    scanf("%d",&t);
    while(t--)
    {
        o++;
        int n;
        ans = 0;
        scanf("%d",&n);
        for (int i = 1; i <= n; i++)
            scanf("%d",&a[i]);
        MergeSort(a,1,n);
        printf("Scenario #%d:
%lld

",o,ans);
    }
    return 0;
}

法三：树状数组 o(nlogn)

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N=2000005;
int c[N],a[N],n;
struct node
{
    int data;
    int pos;
    friend bool operator < (const node a,const node b)
    {
        if(a.data==b.data)
            return a.pos < b.pos;
        else
            return a.data < b.data;
    }
} g[N];
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int data)
{
    while(x <= N)
    {
        c[x]+=data;
        x+=lowbit(x);
    }
}
int sum(int x)
{
    int ans = 0;
    while(x > 0)
    {
        ans+=c[x];
        x-=lowbit(x);
    }
    return ans;
}
int main()
{
    int t,o = 0;
    scanf("%d",&t);
    while(t--)
    {
        o++;
        scanf("%d",&n);
        for (int i = 1; i <= n; i++)
        {
            scanf("%d",&g[i].data);
            g[i].pos = i;
        }
        sort(g+1,g+1+n);
        memset(c,0,sizeof(c));
        for (int i = 1; i <= n; i++)//离散化
            a[g[i].pos] = i;
        long long ans = 0;
        for (int i = 1; i <= n; i++)
        {
            update(a[i],1);
            ans+=i-sum(a[i]);
        }
        printf("Scenario #%d:
%lld

",o,ans);
    }
    return 0;
}